# strictly monotonic

• Jan 24th 2010, 08:13 PM
derek walcott
strictly monotonic
Let K be a subset of R
A function F: K--> R is said to be strictly monotonic if it is either strictly increasing or strictly decreasing. That is, one of the following 2 holds:

1.) x,y are elements in K with x<y implies that f(x)<f(y)
or
2.) x,y are elements in K with x<y implies that f(x)>f(y)

Let I be an interval in R and let f: I-->R be a continuous function.
Prove that f is one-to-one iff f is strictly monotonic.

I am having trouble starting this problem and any help would be much appreciated.
• Jan 24th 2010, 10:08 PM
Drexel28
Quote:

Originally Posted by derek walcott
Let K be a subset of R
A function F: K--> R is said to be strictly monotonic if it is either strictly increasing or strictly decreasing. That is, one of the following 2 holds:

1.) x,y are elements in K with x<y implies that f(x)<f(y)
or
2.) x,y are elements in K with x<y implies that f(x)>f(y)

Let I be an interval in R and let f: I-->R be a continuous function.
Prove that f is one-to-one iff f is strictly monotonic.

I am having trouble starting this problem and any help would be much appreciated.

Suppose that \$\displaystyle f\$ was injective and continuous. Assume for a second that \$\displaystyle f\$ was not monotonic, then what would the IVT give us as a contradiction?

Conversely, if \$\displaystyle x\ne y\$ then either \$\displaystyle x<y\implies f(x)<f(y)\implies f(x)\ne f(y)\$ or \$\displaystyle y<x\implies f(y)<f(x)\implies f(y)\ne f(x)\$, either way we have injectivity.
• Jan 25th 2010, 12:52 PM
derek walcott
so proving that it is an injection proves that it is also one-to-one?
• Jan 25th 2010, 01:03 PM
Drexel28
Quote:

Originally Posted by derek walcott
so proving that it is an injection proves that it is also one-to-one?

Injection means one-to-one.
• Jan 25th 2010, 01:14 PM
derek walcott
okay that's what i thought.

and it's only injective because f is continuous. if f were not continuous then x=y and f(x) ≠ f(y) could occur?
• Jan 25th 2010, 01:19 PM
Drexel28
Quote:

Originally Posted by derek walcott
okay that's what i thought.

and it's only injective because f is continuous. if f were not continuous then x=y and f(x) ≠ f(y) could occur?

Wait, what?
• Jan 25th 2010, 01:30 PM
derek walcott
i am trying to think of the implications of f:K-->R being continuous and not being continuous

how does the problem change? would that prove that it isn't monotonic?
• Jan 25th 2010, 01:39 PM
Drexel28
Quote:

Originally Posted by derek walcott
i am trying to think of the implications of f:K-->R being continuous and not being continuous

how does the problem change? would that prove that it isn't monotonic?

If the function is continuous and injective it is monotonic. If a function is monotonic it is automatically injective.